curve sketching

Been trying to get past this but it just keeps building on each other. It's for tomorrow:

a) Find any vertical and horizontal asys
b) where axis are crossed
c) show curve has no turning points
d) sketch curve, labeling it properly.
e) sketch f(-x) on the graph.

So vertical asympotes are as simple as LCM equal to 0 and substitue. Vertical?!?

Axis cross should be x = 0 and y = 0. It is (0,0) so it should be solved like that.

Curve's turning points. I want to see if differentiating should be enough, or do i need to go down to and make it =0. If the answer is imaginary it's proved, correct?

sketching curve should be done using the information gained from the previous questions and finding out which side the curve goes when approaching the asymptotes (asy+0.0001), (asy - 0.00001) to get a quick estimate.

Can't work the horizontal asymptotes either way. Not sure how you work with and .

do you know what limits are? this is calculus since you are talking about derivatives, so you should know about limits (in fact, the derivative is a limit!). you have a first degree polynomial divided by a second degree one, as x gets larger in either direction, the limit goes to zero. hence, y = 0 is the horizontal asymptote

do you know what limits are? this is calculus since you are talking about derivatives, so you should know about limits (in fact, the derivative is a limit!). you have a first degree polynomial divided by a second degree one, as x gets larger in either direction, the limit goes to zero. hence, y = 0 is the horizontal asymptote

Uhh, I do know about limits, but i can't recall those wicked symbols :